The area of the triangle enclosed by the straight lines $x = 0$,$y = 0$ and $x + 2y + 3 = 0$ in sq. units is:

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively, is:

Suppose $ABCD$ $(AB \parallel CD)$ is a trapezium such that the diagonals $AC$ and $BD$ bisect the angles $\angle DAB$ and $\angle CBA$,respectively. Then

If two opposite vertices of a square are $(5, -4)$ and $(-3, 2)$,find its area.

The area of the triangle formed by the lines $x = 0$,$y = 0$,and $x/a + y/b = 1$ is: